Chapter 7, Problem 7.49P

(a)

To determine

Show that $\overrightarrow{\text{B}}$ can be written as $\overrightarrow{\text{B}}\text{=}\nabla \times \overrightarrow{\text{A}}$ using cartesian and cylindrical coordinate systems.

Answer to Problem 7.49P

It is proved that $\overrightarrow{\text{B}}\text{=}\nabla \times \overrightarrow{\text{A}}$ using cartesian and cylindrical coordinate systems.

Explanation of Solution

Write the expression vector potential.

    $\text{A}=\frac{1}{2}\text{B}\times \text{r}$

Here, $\text{A}$ is the vector potential, $\text{B}$ is the magnetic field and $\text{r}$ is the radial vector.

Take curl of above equation on both sides.

    $\begin{array}{c}\nabla \times \text{A}=\nabla \times \left(\frac{1}{2}\text{B}\times \text{r}\right)\\ =\frac{1}{2}\left[\text{B}\left(\nabla \cdot \text{r}\right)-\left(\text{B}\cdot \nabla \right)\text{r}\right]\end{array}$

Write the expression for radial vector.

    $\text{r}=x\widehat{\text{i}}\text{+}y\widehat{\text{j}}\text{+}z\widehat{\text{k}}$

Take divergence of above equation on both sides.

    $\begin{array}{c}\nabla \cdot \text{r}=\nabla \cdot \left(x\widehat{\text{i}}\text{+}y\widehat{\text{j}}\text{+}z\widehat{\text{k}}\right)\\ \nabla \cdot \text{r}=\left(\widehat{\text{i}}\frac{\partial }{\partial x}+\widehat{\text{j}}\frac{\partial }{\partial y}+\widehat{\text{k}}\frac{\partial }{\partial z}\right)\cdot \left(x\widehat{\text{i}}\text{+}y\widehat{\text{j}}\text{+}z\widehat{\text{k}}\right)\\ =\frac{\partial x}{\partial x}+\frac{\partial y}{\partial y}+\frac{\partial z}{\partial z}\\ =1+1+1\\ =3\end{array}$

Write the general vector form of magnetic field in cartesian coordinate system.

    $\text{B}\text{=}{B}_x\widehat{\text{i}}+B_y\widehat{\text{j}}+B_z\widehat{\text{k}}$

Find the value of $\text{B}\cdot \nabla$.

    $\begin{array}{c}\left(\text{B}\cdot \nabla \right)=\left(B_x\widehat{\text{i}}+B_y\widehat{\text{j}}+B_z\widehat{\text{k}}\right)\cdot \left(\widehat{\text{i}}\frac{\partial }{\partial x}+\widehat{\text{j}}\frac{\partial }{\partial y}+\widehat{\text{k}}\frac{\partial }{\partial z}\right)\\ =B_x\frac{\partial }{\partial x}+B_y\frac{\partial }{\partial y}+B_z\frac{\partial }{\partial z}\end{array}$

Find $\left(\text{B}\cdot \nabla \right)\text{r}$.

    $\begin{array}{c}\left(\text{B}\cdot \nabla \right)\text{r}\text{=}\left(B_x\frac{\partial }{\partial x}+B_y\frac{\partial }{\partial y}+B_z\frac{\partial }{\partial z}\right)\left(x\widehat{\text{i}}\text{+}y\widehat{\text{j}}\text{+}z\widehat{\text{k}}\right)\\ =B_x\frac{\partial }{\partial x}\left(x\widehat{\text{i}}\text{+}y\widehat{\text{j}}\text{+}z\widehat{\text{k}}\right)+B_y\frac{\partial }{\partial y}\left(x\widehat{\text{i}}\text{+}y\widehat{\text{j}}\text{+}z\widehat{\text{k}}\right)+B_z\frac{\partial }{\partial z}\left(x\widehat{\text{i}}\text{+}y\widehat{\text{j}}\text{+}z\widehat{\text{k}}\right)\\ =B_x\widehat{\text{i}}+B_y\widehat{\text{j}}+B_z\widehat{\text{k}}\\ =\text{B}\end{array}$

Conclusion:

Substitute 3 for $\nabla \cdot \text{r}$ and for $\left(\text{B}\cdot \nabla \right)\text{r}$ in the above equation.

    $\begin{array}{c}\nabla \times \text{A=}\frac{1}{2}\left[\text{B}\left(\nabla \cdot \text{r}\right)-\left(\text{B}\cdot \nabla \right)\text{r}\right]\\ =\frac{1}{2}\left[\text{B}\left(3\right)-\text{B}\right]\\ =\frac{1}{2}\left(2\text{B}\right)\\ =\text{B}\end{array}$

Thus, it is proved that $\nabla \times \text{A=B}$ with $\text{A}=\frac{1}{2}\text{B}\times \text{r}$.

Write the expression for $\nabla \times \text{A}$ in cylindrical polar coordinates.

    $\nabla \times \text{A=}\widehat{\rho }\left[\frac{1}{\rho }\frac{\partial }{\partial \varphi }{\text{A}}_z-\frac{\partial }{\partial z}{\text{A}}_{\varphi }\right]+\widehat{\varphi }\left[\frac{\partial }{\partial z}{\text{A}}_{\rho }-\frac{\partial }{\partial \rho }{\text{A}}_z\right]+\widehat{z}\frac{1}{\rho }\left[\frac{\partial }{\partial \rho }\left(\rho A_{{}_{\varphi }}\right)-\frac{\partial }{\partial \varphi }{A}_{\rho }\right]$

Write the expression for $\text{A}$ in terms of $\text{B}$ in cylindrical polar coordinates.

    $\text{A=}\frac{1}{2}B\rho \widehat{\varphi }$

Following values can be obtained using the above equation.

    $\begin{array}{c}{\text{A}}_{\rho }=0\\ {\text{A}}_{\varphi }=\frac{1}{2}B\rho \\ {\text{A}}_z=0\end{array}$

Substitute $0$ for ${\text{A}}_{\rho }$, $\frac{1}{2}B\rho$ for ${\text{A}}_{\varphi }$ and $0$ for ${\text{A}}_z$ in the above equation.

    $\begin{array}{c}\nabla \times \text{A=}\widehat{\rho }\left[\frac{1}{\rho }\frac{\partial }{\partial \varphi }\left(0\right)-\frac{\partial }{\partial z}\left(\frac{1}{2}B\rho \right)\right]+\widehat{\varphi }\left[\frac{\partial }{\partial z}\left(0\right)-\frac{\partial }{\partial \rho }\left(0\right)\right]+\widehat{\text{z}}\frac{1}{\rho }\left[\frac{\partial }{\partial \rho }\left(\rho \frac{1}{2}B\rho \right)-\frac{\partial }{\partial \varphi }\left(0\right)\right]\\ =\widehat{\text{z}}\frac{1}{\rho }\frac{\partial }{\partial \rho }\left(\frac{1}{2}B{\rho }^2\right)\\ =\widehat{\text{z}}\frac{1}{\rho }\frac{2B\rho }{2}\\ =\text{B}\widehat{\text{z}}\end{array}$

Thus, it is proved that $\nabla \times \text{A=B}$ with $\text{A}=\frac{1}{2}B\rho \widehat{\varphi }$.

Thus, it is proved that $\overrightarrow{\text{B}}\text{=}\nabla \times \overrightarrow{\text{A}}$ using cartesian and cylindrical coordinate systems.

(b)

To determine

Give Lagrangian function in cylindrical polar coordinates and corresponding three Lagrange equations.

Answer to Problem 7.49P

Cylindrical form of Lagrangian is $L=\frac{1}{2}m{\left(\dot{\rho }+\rho \dot{\varphi }+\dot{z}\right)}^2+q\left(\dot{\rho }+\rho \dot{\varphi }+\dot{z}\right)\cdot \text{A}$ and the Lagrange equations are $m\ddot{\rho }=m\rho {\dot{\varphi }}^2+qB\rho \dot{\varphi }$, $\frac{d}{dt}\left(m{\rho }^2\dot{\varphi }+\frac{1}{2}qB{\rho }^2\right)=0$ and $m\ddot{z}=0$.

Explanation of Solution

Write the expression for kinetic energy.

    $T=\frac{1}{2}m{\dot{\text{r}}}^2$

Here, $T$ is the kinetic energy, $m$ is the mass of particle and $\dot{\text{r}}$ is the velocity.

Write the expression for potential energy.

    $U=-q\dot{\text{r}}\cdot \text{A}$

Here, $q$ is the charge.

Write the equation for Lagrangian function.

    $L=T-U$

Substitute $\frac{1}{2}m{\dot{\text{r}}}^2$ for $T$ and $-q\dot{\text{r}}\cdot \text{A}$ for $U$ in the above equation.

    $\begin{array}{c}L=\frac{1}{2}m{\dot{\text{r}}}^2-\left(-q\dot{\text{r}}\cdot \text{A}\right)\\ =\frac{1}{2}m{\dot{\text{r}}}^2+q\dot{\text{r}}\cdot \text{A}\end{array}$

Write the equation for $\dot{\text{r}}$ in cylindrical polar coordinates.

    $\dot{\text{r}}=\dot{\rho }+\rho \dot{\varphi }+\dot{z}$

Rewrite the equation for $L$ by substituting the above equation.

    $L=\frac{1}{2}m{\left(\dot{\rho }+\rho \dot{\varphi }+\dot{z}\right)}^2+q\left(\dot{\rho }+\rho \dot{\varphi }+\dot{z}\right)\cdot \text{A}$

Substitute $\frac{1}{2}B\rho \widehat{\varphi }$ for $\text{A}$.

    $\begin{array}{c}L=\frac{1}{2}m{\left(\dot{\rho }+\rho \dot{\varphi }+\dot{z}\right)}^2+q\left(\dot{\rho }+\rho \dot{\varphi }+\dot{z}\right)\cdot \left(\frac{1}{2}B\rho \widehat{\varphi }\right)\\ =\frac{1}{2}m{\left(\dot{\rho }+\rho \dot{\varphi }+\dot{z}\right)}^2+\frac{1}{2}qB{\rho }^2\dot{\varphi }\end{array}$        (I)

Differentiate the above equation with respect to $\rho$.

    $\frac{\partial L}{\partial \rho }=m\rho {\dot{\varphi }}^2+qB\rho \dot{\varphi }$

Differentiate the equation $L=\frac{1}{2}m{\left(\dot{\rho }+\rho \dot{\varphi }+\dot{z}\right)}^2+\frac{1}{2}qB{\rho }^2\dot{\varphi }$ with respect to $\dot{\rho }$.

    $\frac{\partial L}{\partial \dot{\rho }}=m\dot{\rho }$        (II)

Differentiate the above equation with respect to t.

    $\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\rho }}\right)=m\ddot{\rho }$

Differentiate equation (I) with respect to $\varphi$.

    $\frac{\partial L}{\partial \varphi }=0$

Differentiate equation (I) with respect to $\dot{\varphi }$.

    $\frac{\partial L}{\partial \dot{\varphi }}=m{\rho }^2\dot{\varphi }+\frac{1}{2}qB{\rho }^2$        (III)

Differentiate equation (I) with respect to $z$.

    $\frac{\partial L}{\partial z}=0$        (IV)

Differentiate equation (I) with respect to $\dot{z}$.

    $\frac{\partial L}{\partial \dot{z}}=2m\dot{z}$

Write the Lagrangian equation for $\rho$.

    $\begin{array}{c}\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\rho }}\right)=\frac{\partial L}{\partial \rho }\\ \frac{d}{dt}\left(m\dot{\rho }\right)=m\rho {\dot{\varphi }}^2+qB\rho \dot{\varphi }\\ m\ddot{\rho }=m\rho {\dot{\varphi }}^2+qB\rho \dot{\varphi }\end{array}$

Thus, the Lagrange equation for $\rho$ is $m\ddot{\rho }=m\rho {\dot{\varphi }}^2+qB\rho \dot{\varphi }$.

Write the Lagrangian equation for $\varphi$.

  $\begin{array}{c}\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\varphi }}\right)=\frac{\partial L}{\partial \varphi }\\ \end{array}$   

Substitute equation (III) in the above equation.

    $\frac{d}{dt}\left(m{\rho }^2\dot{\varphi }+\frac{1}{2}qB{\rho }^2\right)=0$

Thus, the Lagrange equation for $\varphi$ is obtained as. $\frac{d}{dt}\left(m{\rho }^2\dot{\varphi }+\frac{1}{2}qB{\rho }^2\right)=0$.

Write the Lagrangian equation for z.

    $\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{z}}\right)=\frac{\partial L}{\partial z}$

Substitute equation (IV) in the above equation.

  $\begin{array}{c}\frac{d}{dt}\left(2m\dot{z}\right)=0\\ m\ddot{z}=0\end{array}$

Thus, the cylindrical form of Lagrangian is $L=\frac{1}{2}m{\left(\dot{\rho }+\rho \dot{\varphi }+\dot{z}\right)}^2+q\left(\dot{\rho }+\rho \dot{\varphi }+\dot{z}\right)\cdot \text{A}$ and the Lagrange equations are $m\ddot{\rho }=m\rho {\dot{\varphi }}^2+qB\rho \dot{\varphi }$, $\frac{d}{dt}\left(m{\rho }^2\dot{\varphi }+\frac{1}{2}qB{\rho }^2\right)=0$ and $m\ddot{z}=0$.

(c)

To determine

Explain in detail about the solutions which includes $\rho$ as constant.

Answer to Problem 7.49P

Solutions which includes $\rho$ as constant is detailed.

Explanation of Solution

Since $\rho$ is a constant quantity, its first derivative is zero.

    $\dot{\rho }=0$

Its second derivative will be also zero.

    $\ddot{\rho }=0$

Rewrite the equation $m\ddot{\rho }=m\rho {\dot{\varphi }}^2+qB\rho \dot{\varphi }$ by substituting $0$ for $\ddot{\rho }$.

    $\begin{array}{c}m\left(0\right)=m\rho {\dot{\varphi }}^2+qB\rho \dot{\varphi }\\ 0=\dot{\varphi }\left(m\rho +qB\rho \right)\end{array}$

Solutions of above equation is as follows.

    $\dot{\varphi }=0\Rightarrow \varphi =\text{constant}$

Therefore, as $\varphi$ is a constant, Lagrangian equation for $\varphi$ is valid.